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Mathematics · Differential Equations

JEE Main 2025 — 22 January, Evening Shift — Question 10

If x=f(y)x=f(y) is the solution of the differential equation (1+y2)+(x−2etan⁡−1y)dydx=0,y∈(−π2,π2)\left(1+y^{2}\right)+\left(x-2 e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0, y \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) with f(0)=1f(0)=1, then f(13)f\left(\frac{1}{\sqrt{3}}\right) is equal to :

  1. Option A:

    eπ/4\mathrm{e}^{\pi / 4}

  2. Option B:

    eπ/12e^{\pi / 12}

  3. Option C:

    eπ/3e^{\pi / 3}

  4. Option D:

    eπ/6e^{\pi / 6}

    Correct

Answer: D

Step-by-step solution

We have the differential equation: (1+y2)+(x−2etan⁡−1y)dydx=0.\mathrm{We\ have\ the\ differential\ equation: }\ (1+y^{2})+\left(x-2e^{\tan^{-1}y}\right)\frac{dy}{dx}=0. ⇒dydx=−1+y2x−2etan⁡−1y⇒dxdy=−x−2etan⁡−1y1+y2.\Rightarrow \frac{dy}{dx}=-\frac{1+y^{2}}{x-2e^{\tan^{-1}y}} \quad \Rightarrow \quad \frac{dx}{dy}=-\frac{x-2e^{\tan^{-1}y}}{1+y^{2}}. This is a linear differential equation in x:dxdy+11+y2 x=2etan⁡−1y1+y2.\mathrm{This\ is\ a\ linear\ differential\ equation\ in}\ x:\quad \frac{dx}{dy}+\frac{1}{1+y^{2}}\,x=\frac{2e^{\tan^{-1}y}}{1+y^{2}}. Integrating factor: μ(y)=exp⁡ ⁣(∫11+y2dy)=exp⁡(tan⁡−1y).\mathrm{Integrating\ factor: }\ \mu(y)=\exp\!\left(\int \frac{1}{1+y^{2}}dy\right) =\exp(\tan^{-1}y). Multiplying throughout:ddy ⁣(xetan⁡−1y)=2e2tan⁡−1y1+y2.\mathrm{Multiplying\ throughout: }\quad \frac{d}{dy}\!\Big(x e^{\tan^{-1}y}\Big) =\frac{2e^{2\tan^{-1}y}}{1+y^{2}}.

Let t=tan⁡−1yt=\tan^{-1}y, then dt=dy1+y2dt=\frac{dy}{1+y^{2}}. Hence

xet=∫2e2t dt+C=e2t+C.x e^{t}=\int 2e^{2t}\,dt + C = e^{2t}+C. ⇒x=et+Ce−t=etan⁡−1y+Ce−tan⁡−1y.\Rightarrow \quad x=e^{t}+Ce^{-t}=e^{\tan^{-1}y}+C e^{-\tan^{-1}y}. Applying the condition f(0)=1:1=e0+Ce0⇒C=0.\mathrm{Applying\ the\ condition}\ f(0)=1: \quad 1=e^{0}+C e^{0}\quad\Rightarrow\quad C=0.

Thus,

f(y)=etan⁡−1y.f(y)=e^{\tan^{-1}y}.

Finally,

f ⁣(13)=etan⁡−1(1/3)=eπ/6.f\!\left(\tfrac{1}{\sqrt{3}}\right) =e^{\tan^{-1}(1/\sqrt{3})} =e^{\pi/6}. eπ/6\boxed{e^{\pi/6}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Differential Equations
Topic
Methods of solving a First Order,First Degree Differential

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